full-splitting miller trees and infinitely often equal...

Ros lanowski and Spinas dichotomiesMarczewski-type regularity properties

Full-splitting Miller trees and infinitely often equaltrees

Giorgio Laguzzi(joint with Yurii Khomskii)

Universitat Hamburg

Poznan - September 2014

Giorgio Laguzzi(joint with Yurii Khomskii) Full-splitting Miller trees and infinitely often equal trees


P-dichotomies

Let I be a σ-ideal over the reals, and let P be a forcing with treeconditions.

Definition

We say that a set of reals X satisfies the (I,P)-dichotomy iffeither X ∈ I or there exists T ∈ P such that [T ] ⊆ X .

Well-known examples:

Perfect set property: I = ideal of countable sets / P = Sacksforcing

Kσ-regularity: I = ideal of bounded sets / P = Miller forcing



Such dichotomies provide a dense embedding

P ↪→ Borel \ I.

They are useful because we can use both the combinatorialproperties of trees and the properties of the σ-ideal for studyingthe forcing notion associated. As an example, one can consider thefollowing result.

Theorem (Zapletal)

If I is a σ-ideal on ωω σ-generated by closed sets then the forcingBorel \ I is proper and preserves Baire category (non-meagerground-model sets remain non-meager in the extension).



The perfect set property is related to Davis’ game on the Cantorspace 2ω. The analogous game played on the Baire space ωω givesrise to the following dichotomy.

Definition

We say that a tree T ⊆ ω<ω is full-splitting Miller iff for everysplitting node t ∈ T for all n ∈ ω, tan ∈ T .

Given f : ω<ω → ω, letDf := {x ∈ ωω : ∀∞n(f (x�n) 6= x(n))} and thenDω := {Df : f : ω<ω → ω}.We say that a set X ⊆ ωω satisfies the (Dω,FM)-dichotomy(or Ros lanowski dichotomy) iff either X ∈ Dω or there existsT ∈ FM such that [T ] ⊆ X .

Theorem (Ros lanowski)

Every Σ11 set satisfies the Ros lanowski dichotomy.




Definition


Given f : ω<ω → ω, letDf := {x ∈ ωω : ∀∞n(f (x�n) 6= x(n))} and thenDω := {Df : f : ω<ω → ω}.

We say that a set X ⊆ ωω satisfies the (Dω,FM)-dichotomy(or Ros lanowski dichotomy) iff either X ∈ Dω or there existsT ∈ FM such that [T ] ⊆ X .






Definition


Given f : ω<ω → ω, letDf := {x ∈ ωω : ∀∞n(f (x�n) 6= x(n))} and thenDω := {Df : f : ω<ω → ω}.We say that a set X ⊆ ωω satisfies the (Dω,FM)-dichotomy(or Ros lanowski dichotomy) iff either X ∈ Dω or there existsT ∈ FM such that [T ] ⊆ X .





A slightly different σ-ideal has been studied by Spinas.

Definition

For every x ∈ ωω let Kx := {y ∈ ωω | ∀∞n(x(n) 6= y(n))}, and letIioe be the σ-ideal generated by Kx , for x ∈ ωω.

The two ideals are very similar, and in fact the following equalitieshold.

Proposition

1 cov(Iioe) = cov(Dω) = cov(M)

2 non(Iioe) = non(Dω) = non(M).

3 add(Iioe) = add(Dω) = ω1

4 cof(Iioe) = cof(Dω) = c.



The notion of a full-splitting Miller tree is not sufficient to get theright dichotomy for Iioe, as the following example shows.

Example

Let T be the tree on ω<ω defined as follows:

If |s| is even then SuccT (s) = {0, 1}.If |s| is odd then

SuccT (s) =

{2N if s(|s| − 1) = 02N + 1 if s(|s| − 1) = 1

where SuccT (s) := {n | sa 〈n〉 ∈ T}. Clearly T is Iioe-positivebut cannot contain a full-splitting subtree.



The right dichotomy for Iioe involves a subtle modification of thenotion of a full-splitting Miller tree.

Definition

A tree T ⊆ ωω is called an infinitely often equal tree, or simplyioe-tree, if for each t ∈ T there exists N > |t|, such that for everyk ∈ ω there exists s ∈ T extending t such that s(N) = k . Let IEdenote the partial order of ioe-trees ordered by inclusion.

Definition

We say that a set X ⊆ ωω satisfies the (Iioe, IE)-dichotomy (orSpinas dichotomy) iff either X ∈ Iioe or there exists T ∈ IE suchthat [T ] ⊆ X .

Theorem (Spinas)

Every Σ11 set satisfies the Spinas dichotomy.



Simple remarks about FM and IE

FM adds a Cohen real (let {sn : n ∈ ω} be a fixedenumeration of ω<ω and consider the function ϕ defined byϕ(x) = sx(0)

asx(1)asx(2)

a . . . ).

IE ∗ IE adds a Cohen real.

IE below a certain condition is equivalent to FM. Such acondition is constructed in the following way:

1 If s 6= t are splitting nodes of TGS then |s| 6= |t|.2 If t ∈ TGS is a not the immediate successor of a splitting node

of T then t(|t| − 1) = 0.

Hence, IE also adds Cohen reals.



Dichotomies for higher projective levels

Theorem (Khomskii - L.)

1 Σ12(FM-dich)

2 Σ12(IE-dich)

3 ∀r ∈ ωω (ωL[r ]1 < ω1)


Let κ be inaccessible and let G be Coll(ω,< κ)-generic over V .Then in V [G ] all sets definable from countable sequences ofordinals satisfy the FM- and the IE-dichotomy, and in L(R)V [G ] allsets of reals satisfy the FM- and IE-dichotomy.



Marczewski-type regularity properties



FM and IE-measurability . . .

In the following definitions P is any tree-like forcing notion.

Definition

A set of reals X is said to be P-measurable iff∀T ∈ P∃T ′ ∈ P,T ′ ⊆ T (X ∩ [T ′] = ∅ ∨ [T ′] ⊆ X ).

A set of reals X is said to be weakly-P-measurable iff∃T ∈ P(X ∩ [T ] = ∅ ∨ [T ] ⊆ X ).

Such a notion of measurability generalizes the Lebesguemeasurability (P = random forcing), the Bernstein partitionproperty (P =Sacks forcing) and the Ramsey property (P =Mathias forcing).

Theorem (Brendle - Lowe)

Let Γ be a topologically reasonable family of sets of reals. Then formany tree forcings P one has Γ(P)⇔ Γ(wP).



. . . vs Baire property

Question. How are FM- and IE-measurability related to the othernotions of regularity?


Let Γ be a pointclass closed under continuous pre-images. Thenthe following are equivalent:

1 Γ(Baire)

2 Γ(FM)

3 Γ(IE)



Proof.

2⇒ 3. We say that an ioe-tree T is in strict form if it can bewritten as follows:

for every σ ∈ ω<ω, there exists Nσ ⊆ ωn, for some n ≥ 1, suchthat∀k ∃!s ∈ Nσ (s(n − 1) = k), andfor m < (n − 1), there is some k such that s(m) 6= k for alls ∈ Nσ. We use len(Nσ) = n to denote the length of Nσ, andwe canonically enumerate Nσ as {sσk | k < ω}, in such a waythat sσk (n − 1) = k .T is the tree generated by sequences of the form

s∅n0a s〈n0〉

n1a s〈n0,n1〉

n2a . . . a s

〈n0,n1,n2,...,n`−1〉n`

for some sequence 〈n0, n1, . . . , n`〉.



. . . .

If T is in strict form, t ∈ T , and we need to find the first N suchthat ∀k ∃s ⊇ t with s(N) = k , we only need to find the firstsequence of the form sσn extending t, and thenN := |sσn |+ len(Nσa〈n〉). Every ioe-tree T can be pruned to anioe-subtree S ≤ T in strict form.So, let A ∈ Γ and let T be an ioe-tree in strict form. Define afunction ψ′ : ω<ω → T by setting:

ψ′(〈x(0), . . . , x(`)〉) := s∅x(0) a s〈x(0)〉x(1) a s

〈x(0),x(1)〉x(2) a . . . a sx�`x(`).

This gives rise to a natural homeomorphism ψ : ωω ∼= [T ]. Sinceψ−1[A] is also in Γ we can find a full-splitting tree S such that[S ] ⊆ ψ−1[A] or [S ] ∩ ψ−1[A] = ∅. But ψ“[S ] generates anioe-subtree of T ; in fact, if t is a splitting node of S then for everyn we have ψ′(t a 〈n〉) = ψ′(t) a stn, and the last digit of stn is n bydefinition.



. . . .

3⇒ 2. Let A ∈ Γ and recall the Golstern-Shelah tree TGS

introduced before. Since A is IE-measurable there existsS ≤ TGS such that [S ] ⊆ A or [S ] ∩ A = ∅. But then S is anFM-tree , so A is weakly FM-measurable, which is sufficient.



. . . .

1⇒ 2. Let A ⊆ ωω be a set in Γ. By Γ(Baire) we can find abasic open set [s] such that [s] ⊆∗ A or [s] ∩ A =∗ ∅, where⊆∗ and =∗ stands for “modulo a meager set”. Without lossof generality, assume the former. Then there is a Gδco-meager set B ⊆ [s] ∩ A. Since Dω ⊆M, B cannot beDω-small, hence it contains an FM-tree. This is sufficient, asFM- and weak FM-measurability are classwise equivalent.

2⇒ 1. Use the function ϕ described before. Let A ∈ Γ andlet A′ := ϕ−1[A], also in Γ. By Γ(FM) we can find a T ∈ FMsuch that [T ] ⊆ A′ or [T ] ∩ A′ = ∅, without loss of generalitythe former. Then ϕ“[T ] cannot be meager, but it is analytic,so it is comeager in some [s]. Then [s] ⊆∗ A. This is sufficientbecause Γ(Baire) is equivalent to the assertion that for allA ∈ Γ there exists a basic open [s] such that [s] ⊆∗ A or[s] ∩ A =∗ ∅.



Σ12(wIE) vs Σ1

2(IE)


∆12(Baire)⇒ Σ1

2(wIE).

Corollary

It is consistent that Σ12(wIE) is true while Σ1

2(IE) is false; inparticular IE- and weak IE-measurability are not classwiseequivalent.



Proof of Theorem.

Assume ∆12(Baire). Let A ⊆ ωω be a Σ1

2 set. We have to find anIE-tree T such that [T ] ⊆ A or [T ] ∩ A = ∅.

We may assume that for some r , ωL[r ]1 = ω1, since otherwise

Σ12(wIE) follows easily (for example from Σ1

2(Baire)). We mayalso assume, without loss of generality, that the parameters in thedefinition of A are in L[r ]. Using the Borel decomposition of Σ1

2

sets we can write A =⋃α<ω1

Bα, where Bα are Borel sets coded inL[r ]. If there exists at least one α such that Bα /∈ Iioe, then thereis an IE-tree T with [T ] ⊆ Bα ⊆ A and we are done. So supposethat all Bα are Iioe-small. For each α, since L[r ] |=“Bα ∈ Iioe” wecan fix a sequence 〈xαi | i < ω〉 of reals in L[r ] such thatBα ⊆

⋃i<ω Kxαi

.



. . . .

Let ρ : ωω → ωω be defined by ρ(x) := 〈x(0), x(2), x(4), . . . 〉. By∆1

2(Baire), we know that in V there is a Cohen real c over L[r ].Then c is infinitely often equal over L[r ], and in particular,infinitely often equal to ρ(xαi ) for all α < ω1, i < ω. Let Tc be theFM-tree such that

[Tc ] = {y | ρ(y) = c}.

We claim that [Tc ] ∩ A = ∅. Let a ∈ A, then there is someα < ω1 such that a ∈ Bα. By absoluteness of “Bα ⊆

⋃i<ω Kxαi

”,there is some i < ω such that a is eventually different from xαi . LetN ∈ ω be such that ∀n > N (a(n) 6= xαi (n)). But since c is ioe toρ(xαi ), we can easily find n > N such that c(n) = xαi (2n) 6= a(2n).By definition this implies that a /∈ [Tc ].



Open questions



A couple of open questions

1 How can we characterize ∆12(wIE) and Σ1

2(wIE) in terms oftrascendental principle over L?

2 Investigate the ideal σ-generated by the sets X satisfying∀T ∈ P∃T ′ ∈ P(T ′ ≤ T ∧ [T ′] ∩ X = ∅), whereP ∈ {FM, IE}.(Work in progress, joint with Yurii Khomskii and WolfgangWohofsky.)

3 Investigate analogous Ros lanowski and Spinas dichotomies inκκ. (Related to Philipp Schlicht’s work)



Thank you for your attention!



A game for IE-dichotomy

Definition

Let G IE(A) be the game in which players I and II play as follows:

I: N0 (s0,N1) (s1,N2) . . .

II: k0 k1 k2 . . .

where si ∈ ω<ω \ {∅}, Ni ≥ 1, ki ∈ ω, and the following rulesmust be obeyed for all i :

|si | = Ni ,

si (Ni − 1) = ki .

Then player I wins iff z := s0as1 a s2

a · · · ∈ A.




1 Player I has a winning strategy in G IE(A) iff there is anIE-tree T such that [T ] ⊆ A.

2 Player II has a winning strategy in G IE(A) iff A ∈ Iioe.


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